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A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute tab indent temperature is is 60 degree in cooling from 60 degree Celsius 50 degree it will take what is the answer options 1) 30 sec 2) 60sec 3) 90sec 4) 50sec

A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute tab indent temperature is is 60 degree in cooling from 60 degree Celsius 50 degree it will take what is the answer options                                   1) 30 sec 
  2) 60sec 
  3) 90sec 
  4) 50sec

Grade:12

2 Answers

Vikas TU
14149 Points
4 years ago
Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] 
( Tf - Ti)/t = K[ ( 80 + 60)/2 - 60]
( 80-60)/1 = K[ 70 - 60]
20 = K× 10 
K = 20/10 = 2
in second condition, 
initial temperature ( Ti) = 60°C 
Final temperature ( Tf) = 50°C 
Time taken for cooling is t 
A/C Newton's law of cooling 
( 60 - 50)/t = 2 [ (60+50)/2 -60]
10/t = 2* 5 
10/t = 10  
t = 1 min.
Khimraj
3007 Points
4 years ago
According to Rate of cooling
dT/dt  = K((T_initial+T_final)/2 - T_surr) 
( 80-60)/1 = K(70-60)
K = 2
Time taken for cooling
(60 - 50)/t = 2 ((60+50)/2 -60)
t = 60 sec

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